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[SOLVED] Quantization noise in ADC

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The_pearl

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Hi,
I found this equation for quantization noise in Analog Integrated Circuit Design by Tony Chan Carusone, David Johns and Kenneth Martin chapter 15 page 610 under the topic Quantization noise using deterministic approach.
Could someone please explain to me how they substitute Vlsb for Vq in the first step ?

1720663809687.png


math.png
 

Solution
As said by wwfeldman, from write upper graph Vq goes from 1/2Vlsb to -1/2Vlsb while the time "t" goes from -T/2 to T/2. It's clearly a ramp so you can describe it by a linear expression:

Vq = a*t + b

sustituting:

1/2*Vlsb = a*(-T/2) + b
-1/2*Vlsb = a*(T/2) + b

solving this simple 2 simultaneous equation we have:

a = -Vlsb/T
b = 0

then

Vq = -Vlsb*t/T
in the upper right graph, Vq vs time, the graph lies between -1/2 Vlsb and +1/2 Vlsb
Vq is measured in Vlsb

Vq is theoretical while Vlsb is practical representation thereof

i would think the explanation is in the text
 

As said by wwfeldman, from write upper graph Vq goes from 1/2Vlsb to -1/2Vlsb while the time "t" goes from -T/2 to T/2. It's clearly a ramp so you can describe it by a linear expression:

Vq = a*t + b

sustituting:

1/2*Vlsb = a*(-T/2) + b
-1/2*Vlsb = a*(T/2) + b

solving this simple 2 simultaneous equation we have:

a = -Vlsb/T
b = 0

then

Vq = -Vlsb*t/T
 

Solution
As said by wwfeldman, from write upper graph Vq goes from 1/2Vlsb to -1/2Vlsb while the time "t" goes from -T/2 to T/2. It's clearly a ramp so you can describe it by a linear expression:

Vq = a*t + b

sustituting:

1/2*Vlsb = a*(-T/2) + b
-1/2*Vlsb = a*(T/2) + b

solving this simple 2 simultaneous equation we have:

a = -Vlsb/T
b = 0

then

Vq = -Vlsb*t/T
Thank you for your explanation I understand now.
 

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